240. Search a 2D Matrix II
题目描述：
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom. For example,
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.
Given target = 20, return false.
题目翻译
写一个高效的算法，从一个m×n的整数矩阵中查找出给定的值，矩阵具有如下特点： 每一行从左到右递增。
每一列从上到下递增。
解题方案
标签： 查找
思路： 从右上角开始, 比较target 和 matrix[i][j]的值. 如果小于target, 则该行不可能有此数, 所以i++; 如果大于target, 则该列不可能有此数, 所以j--. 遇到边界则表明该矩阵不含target.
代码：
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size()==0 || matrix[0].size()==0) return false;
int i=0, j=matrix[0].size()-1;
while(i<matrix.size() && j>=0) {
int x = matrix[i][j];
if(target == x) return true;
else if(target < x) --j;
else ++i;
}
return false;
}
};